[leetCode]Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

字符匹配问题，值得注意的有一点：对一个字符从小到大找过没有匹配之后还要从大到小找一次，如果还是没有匹配才是真的没有匹配

testcase string s = "aaaaaaa", set<String> :{"aaa","aaaa"}就是这样的例子。

#include <string>
#include <set>
#include <algorithm>
using namespace std;
class Solution {
public:    
    bool wordBreak(string s, unordered_set<string> &dict) {
        unordered_set<string>::iterator iter;
        string substr1,substr2;
        substr1 = substr2 = s;
        //short->long
        for(int i = 1; i<= s.size();i++){
            substr1 = s.substr(0,i); 
            iter = dict.find(substr1);
            while(iter == dict.end()&& i < s.size()){
                substr1 = s.substr(0,++i);
                iter = dict.find(substr1);
            }
            if(iter == dict.end() && i == s.size())    break;
            s = s.substr(i);
            i = 0;
        }
        if(s.size() == 0)    return true;
        //long->short
        s = substr2;
        for(int i = s.size(); i > 0;i--){
            substr2 = s.substr(0,i); 
            iter = dict.find(substr2);
            while(iter == dict.end()&& i > 0){
                substr2 = s.substr(0,--i);
                iter = dict.find(substr2);
            }
            if(iter == dict.end() && i == 0)    break;
            s = s.substr(i);
            i = s.size()+1;
        }
        if(s.size() == 0)    return true;
        return false;
    }
};